Proof that limit of convergent sequence lies in $[a,\infty)$

Question

I have a sequence $(a_n)_{n=1}^\infty$, which converges, and has $a_n \in [a,\infty)$. I need to prove that the limit of $a_n$ lies in $[a,\infty)$.

I thought that doing this by contradiction would be the best way to go. I said that $\lim_{n\rightarrow \infty}a_n = L$. $L$ should be in $[a,\infty)$ (with all the other elements of $a_n$).

For the contradiction, I'm saying that $L \notin [a,\infty)$. If this is the case, then $L \in (-\infty, a)$

So I figured to get the contradiction, I would use this 'fact' about L, and then find that the sequence does not converge if that's the case, but I'm not sure how to do that. I've tried using the $\epsilon - N$ definition of a limit but that didn't really get me anywhere.

Thanks for your time

Answer 1

Yes, that's the way to go. Take $\varepsilon=a-L$. This makes sense, because $a-L>0$. Now, let $N\in\mathbb N$. Then there's a $n\geqslant N$ such that $|a_n-L|\geqslant\varepsilon$. Indeed, you can take $n=N$, because$$L<a\leqslant a_n\implies|a_n-L|=a_n-L=\overbrace{a_n-a}^{\phantom{0}\geqslant0}+a-L\geqslant a-L=\varepsilon.$$

Answer 2

$$ L <a \implies a-L =\epsilon >0 $$

Consider $$ B= (L-\epsilon /2,L+\epsilon /2 ) $$

No term of your sequence are included in $B$

That contradicts the fact that $ a_n \to L$